[刷题]算法竞赛入门经典(第2版) 6-4/UVa439 6-5/UVa1600

比较忙比较累，只贴代码了。

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题目：6-4 UVa439 - Knight Moves

//UVa439 - Knight Moves//Accepted 0.000s//#define _XIENAOBAN_#include<iostream>#include<cstring>#include<queue>#define M(po) Map[po.x][po.y]using namespace std;struct poi {    int x, y, weight;    poi operator +(const poi &that) const {        return poi{ x + that.x, y + that.y, weight};    }    bool operator ==(const poi &that) const {        return (x == that.x) && (y == that.y);    }} op, ed;const poi dir[8] = { { 2,1 },{ -2,1 },{ 2,-1 },{ -2,-1 },{ 1,2 },{ -1,2 },{ 1,-2 },{ -1,-2 } };bool Map[10][10];char xstart, ystart, xend, yend;int xs, ys, xe, ye;int BFS(){    if (op == ed) return 0;    queue<poi> Q;    Q.push(op);    M(op) = true;    while (!Q.empty()) {        for (int i(0);i < 8;++i){            poi nxt(Q.front() + dir[i]);            if (nxt.x > 0 && nxt.y > 0 && nxt.x < 9 && nxt.y < 9 && !M(nxt)) {                ++nxt.weight;                if (nxt == ed) return nxt.weight;                M(nxt) = true;                Q.push(nxt);            }        }        Q.pop();    }    return -1;}int main(){#ifdef _XIENAOBAN_#define gets(T) gets_s(T, 129)    freopen("in.txt", "r", stdin);    //freopen("out.txt", "w", stdout);#endif    while (scanf("%c%c %c%c", &xstart, &ystart, &xend, &yend) == 4) {        memset(Map, 0, sizeof(Map));        op.x = xstart - 96, op.y = ystart - 48, op.weight = 0;        ed.x = xend - 96, ed.y = yend - 48, ed.weight = 0;        printf("To get from %c%c to %c%c takes %d knight moves.\n", xstart, ystart, xend, yend, BFS());        while (getchar() != '\n');    }    return 0;}

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题目：6-5 UVa1600 - Patrol Robot

//UVa1600 - Patrol Robot//Accepted 0.000s//#define _XIENAOBAN_#include<iostream>#include<cstring>#include<queue>#define DONE 2333333using namespace std;struct step { int x, y, k, mov; };int T, m, n, k;int Map[24][24],Obst[24][24];void judge(queue<step> &Q, step &now, int x, int y) {    if (Obst[x += now.x][y += now.y] == DONE) return;    int _k = (Obst[x][y] ? now.k + 1 : 0);    if (_k <= k) {        if (_k) {            if (Map[x][y] && Map[x][y] <= _k) return;            Map[x][y] = _k;        }        else Obst[x][y] = DONE;        Q.push(step{ x, y, _k, now.mov + 1 });    }}int main(){#ifdef _XIENAOBAN_#define gets(T) gets_s(T, 129)    freopen("in.txt", "r", stdin);    freopen("out.txt", "w", stdout);#endif    scanf("%d", &T);    while (T--) {        scanf("%d%d%d", &m, &n, &k);        for (int i(1);i <= m;++i) for (int j(1);j <= n;++j)            scanf("%d", &Obst[i][j]);        memset(Map, 0, sizeof(Map));        queue<step> Q;        Q.push(step{ 1,1,0,0 });        Obst[1][1] = DONE;        while (!Q.empty()) {            step &now(Q.front());            if (now.x == m && now.y == n) break;            if (now.x + 1 <= m) judge(Q, now, 1, 0);            if (now.y + 1 <= n) judge(Q, now, 0, 1);            if (now.x - 1 >= 1) judge(Q, now, -1, 0);            if (now.y - 1 >= 1) judge(Q, now, 0, -1);            Q.pop();        }        if (Q.empty()) printf("-1\n");        else printf("%d\n", Q.front().mov);    }    return 0;}